Đáp án:
\({V_{{H_2}}} = 3,36{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
Ta có:
\({n_{Fe}} = \frac{{8,4}}{{56}} = 0,15{\text{ mol = }}{{\text{n}}_{{H_2}}}\)
\( \to {V_{{H_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\)
\({n_{HCl}} = 2{n_{Fe}} = 0,3{\text{ mol}} \to {{\text{m}}_{HCl}} = 0,3.36,5 = 10,95{\text{ gam}} \)
\(\to {{\text{m}}_{dd\;{\text{HCl}}}} = \frac{{10,95}}{{10,95\% }} = 100{\text{ gam}}\)
BTKL:
\({m_{dd}} = {m_{Fe}} + {m_{dd{\text{ HCl}}}} - {m_{{H_2}}} = 8,4 + 100 - 0,15.2 = 108,1{\text{ gam}}\)
\({m_{FeC{l_2}}} = 0,15.(56 + 35,5.2) = 19,05{\text{ gam}}\)
\( \to C{\% _{FeC{l_2}}} = \frac{{19,05}}{{108,1}} = 17,62\% \)
